Solved – Repeated measures: aggregate, fixed effects, or random effects

mixed modelrrepeated measures

I have a repeated measures design with data collected across 8 treatments, with 5-6 repeats for each treatment in closely spaced time intervals (time is not important in the study). I only have 4 subjects in the study, and one dropped out two-thirds of the way through. As far as I can see, I have three options to analyze these data:

1) use a mixed model with subject as a random effect. The main problem here is that there are only 4 subjects and so estimates of the variance of the random effects will be imprecise.

2) treat subject as a fixed effect. Someone suggested this to me, but I'm unclear how this accounts for the fact that the repeated measurements are not independent. Surely it doesn't?

3) collapse the data across the repeated measures so that the response variable is either the sum or mean of all repeated values for a given subject/treatment combination. This obviates the need for a repeated measures design, but it seems I lose some power.

With only 4 subjects, I'm inclined to go with option 3, but I'd love to hear people's thoughts.

Best Answer

I think you should go with subject as a fixed effect since you only have 3 subjects.

What is the goal of the study? If you want to test whether all 8 treatments are the same - or test particular contrasts - you could treat subject like a fixed effect block. If you want to predict the response of an as-yet-unseen subject, you need random effects. You also need a bigger sample.

RE: "dependence of observations." This doesn't matter. Whether subject is treated as random or fixed, you still have repeated measures - i.e. multiple observations nested within subject.

I would go with options 2 and 3, averaging response and treating subject as a fixed blocking factor. You don't lose power in the comparison between treatments because treatments are applied at the "subject" level, not the replicate level. By taking the mean of your repeated measures, you reduce the variance of your response, so you gain that way (i.e. N is smaller but so is $\sigma^2$)

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